Question

find diffrentiation of cos(x^2+bx+c)​

Answer 1

Question

Find the derivative of cos(x² + bx + c)

Solution

Let y = cos(x² + bx + c)

To finD : Derivative of y

Now,

$\tt{y' = \dfrac{d[cos( {x}^{2} + bx + c)] }{dx} } \\ \\ \longrightarrow \: \tt{y' = - sin( {x}^{2} + bx + c). \frac{d( {x}^{2} + bx + c) }{dx} } \\ \\ \longrightarrow \: \tt{y' = [ - sin( {x}^{2} + bx + c)].(2x + b)} \\ \\ \longrightarrow \: \boxed{ \boxed{\tt{y' = - 2x \: sin( {x}^{2} + bx + c) - b \: sin( {x}^{2} + bx + c)}}}$

Answer 2

❏ QuesTion :-

◗ Find the differentiation of cos ( x² + bx + c)

❏ Answer :-

$\to \boxed{ \sf \frac{dy}{dx} =- (2x + b) \: \sin( {x}^{2} + bx + c) \: }$

❏ SoluTion :-

Let ,

$\to \sf y \: = \cos ( {x}^{2} + bx + c)</p><p> \\ \\ \bf differentiate \: with \: respect \: to \: x \\ \\ \to \sf \frac{dy}{dx} = -\sin( {x}^{2} + bx + c) \frac{d}{dx} ( {x}^{2} + bx + c) \\ \\ \to \sf \frac{dy}{dx} = -\sin( {x}^{2} + bx + c) \times (2x + b) \\ \\ \to \boxed{\sf \frac{dy}{dx} = -(2x + b) \: \sin( {x}^{2} + bx + c)}$

Hope it will help you .

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